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Question Number 93963 by  M±th+et+s last updated on 16/May/20

we have for quadratic equations x=((−b±(√(b^2 −4ac)))/(2a)) what about cubic equation is there any rules or ways to solve?

$${we}\:{have}\:{for}\:{quadratic}\:{equations} \\ $$$${x}=\frac{−{b}\pm\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}}{\mathrm{2}{a}} \\ $$$${what}\:{about}\:{cubic}\:{equation}\:{is}\:{there}\:{any} \\ $$$${rules}\:{or}\:{ways}\:{to}\:{solve}? \\ $$

Commented by Kunal12588 last updated on 16/May/20

for the roots I heard there is a formula for cubic too but it is too long to be remembered. same for biquadratic.[don′t know for higher] but there are methods to solve like cardano and trigonometric [I heard of Farrari too, I think it′s for biquadratic]

$${for}\:{the}\:{roots}\:{I}\:{heard}\:{there}\:{is}\:{a}\:{formula}\:{for} \\ $$$${cubic}\:{too}\:{but}\:{it}\:{is}\:{too}\:{long}\:{to}\:{be}\:{remembered}. \\ $$$${same}\:{for}\:{biquadratic}.\left[{don}'{t}\:{know}\:{for}\:{higher}\right] \\ $$$${but}\:{there}\:{are}\:{methods}\:{to}\:{solve}\:{like}\:{cardano}\:{and} \\ $$$${trigonometric}\:\left[{I}\:{heard}\:{of}\:{Farrari}\:{too},\:{I}\:{think}\right. \\ $$$$\left.{it}'{s}\:{for}\:{biquadratic}\right] \\ $$

Commented by mr W last updated on 16/May/20

see Q89687

$${see}\:{Q}\mathrm{89687} \\ $$

Commented by mr W last updated on 16/May/20

once i made following summary for cubic equation:

$${once}\:{i}\:{made}\:{following}\:{summary}\:{for} \\ $$$${cubic}\:{equation}: \\ $$

Commented by mr W last updated on 16/May/20

Commented by  M±th+et+s last updated on 16/May/20

thank you sir

$${thank}\:{you}\:{sir} \\ $$

Commented by Tony Lin last updated on 16/May/20

Commented by Tony Lin last updated on 16/May/20

Commented by Tony Lin last updated on 16/May/20

Commented by Tony Lin last updated on 16/May/20

we still can get the formula of cubic and biquadratic in terms of a,b,c,d,e but they are too hard to remember , which is totally not useful

$${we}\:{still}\:{can}\:{get}\:{the}\:{formula}\:{of}\:{cubic} \\ $$$${and}\:{biquadratic}\:{in}\:{terms}\:{of}\:{a},{b},{c},{d},{e} \\ $$$${but}\:{they}\:{are}\:{too}\:{hard}\:{to}\:{remember}\:, \\ $$$${which}\:{is}\:{totally}\:{not}\:{useful} \\ $$

Commented by  M±th+et+s last updated on 16/May/20

thank you sir

$${thank}\:{you}\:{sir}\: \\ $$

Commented by  M±th+et+s last updated on 16/May/20

sir MR.W can you put an example for this rule and thank you again

$${sir}\:{MR}.{W}\:{can}\:{you}\:{put}\:{an}\:{example}\:{for}\:{this}\:{rule} \\ $$$${and}\:{thank}\:{you}\:{again} \\ $$

Commented by mr W last updated on 16/May/20

you can apply the formulae directly for any cubic equation, i don′t think you need an example. but i give an example: x^3 −3x−4=0 p=−1, q=−2 p^3 +q^2 =−1+4=3>0 ⇒one real root x=(((√3)+2))^(1/3) −(((√3)−2))^(1/3) ≈2.1958

$${you}\:{can}\:{apply}\:{the}\:{formulae}\:{directly}\: \\ $$$${for}\:{any}\:{cubic}\:{equation},\:{i}\:{don}'{t}\:{think} \\ $$$${you}\:{need}\:{an}\:{example}.\:{but}\:{i}\:{give}\:{an} \\ $$$${example}: \\ $$$${x}^{\mathrm{3}} −\mathrm{3}{x}−\mathrm{4}=\mathrm{0} \\ $$$${p}=−\mathrm{1},\:{q}=−\mathrm{2} \\ $$$${p}^{\mathrm{3}} +{q}^{\mathrm{2}} =−\mathrm{1}+\mathrm{4}=\mathrm{3}>\mathrm{0}\:\Rightarrow{one}\:{real}\:{root} \\ $$$${x}=\sqrt[{\mathrm{3}}]{\sqrt{\mathrm{3}}+\mathrm{2}}−\sqrt[{\mathrm{3}}]{\sqrt{\mathrm{3}}−\mathrm{2}}\approx\mathrm{2}.\mathrm{1958} \\ $$

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